# Spherical Polar co-ordinates

Figure 1:- Relationship between spherical polar coordinates and the rectangular Cartesian coordinates.

The spherical polar coordinates represent the coordinates of points on the surface of a sphere in a co-variant form. The coordinates of the point $P$ in this system is represented by the radial vector $r$ which is the distance from the origin to the point, the polar or zenith angle $\theta$ which is the angle the radial vector makes with respect to the z axis and the azimuth or longitudinal angle $\phi$ which is which is the normal polar coordinate in the x − y plane as shown below in the figure.

These co-ordinates are related to the rectangular coordinates x, y, and z through
$$x=r\sin \theta cos\varphi$$
$$y=r\sin \theta \sin \varphi$$
$$z=r\cos \theta$$
Spherical polar coordinates in terms of Cartesian coordinates are
$$r=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$$
$$\tan \theta =\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}}{z}$$
$$\tan \varphi =\frac{y}{z}$$
Unit vectors in spherical coordinate system
1. $\hat {e_r}$ : points towards the $r$ axis that is in the direction of vector $\vec r$ along which only coordinate $r$ changes. We can also write$\vec r= r \hat {e_r}$.
2. $\hat {e_\theta}$: Unit vector $\hat {e_\theta}$ is tangent at $P$ to circle $SPT$ . Displacement along this circle only changes coordinate.
3. $\hat {e_\varphi}$: This unit vector is also tangent at $P$ if circle under consideration is $PP’$ produced by rotation of $OP$ along z-axis. Displacement along this circle only changes $\varphi$.
The general differential displacement for any particle P in spherical polar coordinate is
$$d\vec{r}=dr\widehat{{{e}_{r}}}+rd\theta \widehat{{{e}_{\theta }}}+rsin\theta d\varphi \widehat{{{e}_{\varphi }}}$$
The unit vectors $\hat {e_r}$ ,$\hat {e_\theta}$ and $\hat {e_\varphi}$ can be expressed in terms of $\hat i$ , $\hat j$ and $\hat k$ :
$$\hat {{e_r}} = \sin \theta cos\varphi \hat i + \sin \theta sin\varphi \hat j + cos\theta \hat k$$
$$\hat {{e_\theta }} = \cos \theta cos\varphi \hat i + \cos \theta sin\varphi \hat j – \sin \theta \hat k$$
$$\hat {{e_\varphi }} = – \sin \varphi \hat i + \cos \varphi \hat j$$
The unit vectors $\left( \widehat{{{e}_{r}}},\widehat{{{e}_{\theta }}},\widehat{{{e}_{\varphi }}} \right)$ , unlike $\left( \hat{i},\hat{j},\hat{k} \right)$ are not constant vectors but change in direction as co-ordinates $\theta$ and $\varphi$ change. At each point they constitute an orthogonal right handed co-ordinate system, that is we have
$${{\hat{e}}_{r}}\cdot {{\hat{e}}_{\varphi }}={{\hat{e}}_{r}}\cdot {{\hat{e}}_{\theta }}={{\hat{e}}_{\theta }}\cdot {{\hat{e}}_{\varphi }}=0$$
${{\hat{e}}_{r}}\times {{\hat{e}}_{\theta }}={{\hat{e}}_{\varphi }}$
${{\hat{e}}_{\theta }}\times {{\hat{e}}_{\varphi }}={{\hat{e}}_{r}}$
${{\hat{e}}_{\varphi }}\times {{\hat{e}}_{r}}={{\hat{e}}_{\theta }}$

## Solved Problems on Spherical Polar co-ordinates

Question 1
How to Find Kinetic energy in terms of $(r,\theta ,\phi )$
Solution 1
We have to find the kinetic energy in terms of $(r,\theta ,\phi )$ that is in terms of spherical c0-ordinates. Kinetic energy in terms of Cartesian co-ordinates is

$T = \frac{1}{2}m\left( {{{\dot x}^2} + {{\dot y}^2} + {{\dot z}^2}} \right)$ (1)

Where $\dot x,\dot y{\rm{ and }}\dot z$ are derivatives of z, y and z with respect to time.

Cartesian co-ordinates x, y, z in terms of $r,\theta$ and $\phi$ are

$\begin{array}{l}x = r\sin \theta \cos \phi \\y = r\sin \theta \sin \phi \\z = r\cos \theta \end{array}$

Now derivatives of x, y and z w.r.t. t are

$\begin{array}{l}\dot x = \frac{{dx}}{{dt}} = \dot r\sin \theta \cos \phi + r\cos \theta \cos \phi \dot \theta – r\sin \theta \sin \phi \dot \phi \\\dot y = \frac{{dy}}{{dt}} = \dot r\sin \theta \sin \phi + r\cos \theta \sin \phi \dot \theta + r\sin \theta \cos \phi \dot \phi \\\dot z = \frac{{dz}}{{dt}} = \dot r\cos \theta – r\sin \theta \dot \theta \end{array}$

Where

$\dot r = \frac{{dr}}{{dt}},\dot \theta = \frac{{d\theta }}{{dt}},\dot \phi = \frac{{d\phi }}{{dt}}$

means all $r,\theta$ and $\phi$ changes with time as the particle moves or changes its position with time.

Now calculate for

${\dot x^2},{\dot y^2},{\dot z^2}$

${(\dot x)^2} + {(\dot y)^2} + {(\dot z)^2} = {\dot r^2} + {r^2}{\dot \theta ^2} + {r^2}{\sin ^2}\theta {\dot \phi ^2}$

Putting this value of ${(\dot x)^2} + {(\dot y)^2} + {(\dot z)^2}$in equation 1 we get kinetic energy of particle or system in terms of $r,\theta$ and $\phi$.
Hence

$T = \frac{1}{2}m({\dot r^2} + {r^2}{\dot \theta ^2} + {r^2}{\sin ^2}\theta {\dot \phi ^2})$

## Motion in two dimensions

Let us denote the two dimensional plane as XY Plane as shown below in the figure.

Figure 2:- Motion of the particle in two dimensional plane.

The position vector of the particle is given by
$\vec{r}=x\hat{i}+y\hat{j}$
as $Z=0$ for particle in motion.

In two dimension case the xy plane corresponds to the situation where $\vec{r}=\rho$. The polar angle $\theta$ given in figure 1 is then $\frac{\pi}{2}$ for the particle and spherical polar coordinates $(r, \theta, \phi)$  then reduced to circular polar coordinates $(r, \phi);\theta=90^0$.

Now position of the particle P in terms of radial distance $r$ from origin O and $\phi$ is given by

$r=\sqrt{x^2+y^2}$
$x=rcos\varphi$
$tan\varphi=\frac{x}{y}$
$y=rsin\varphi$

We already have displacement vector $d\vec{r}$ in three dimensionn that is

$d\vec{r}=dr\widehat{{{e}_{r}}}+rd\theta \widehat{{{e}_{\theta }}}+rsin\theta d\varphi \widehat{{{e}_{\varphi }}}$

The displacement vector $d\vec r$ in circular polar coordinates can be obtained by putting $\thata = 90^0$ and $d\thata = 0$ in above equation. So , we get

$d\vec{r}=dr\widehat{{{e}_{r}}}+rd\varphi \widehat{{{e}_{\varphi }}}$

where unit vectors $\hat {{e_r}}$ and $\hat {{e_\varphi }}$ are given by

$$\hat {{e_r}} = cos\varphi \hat i + sin\varphi \hat j$$
$$\hat {{e_\varphi }} = – \sin \varphi \hat i + \cos \varphi \hat j$$

##### Reference material for further reading

1. Refer this link to determine the spherical unit vectors in terms of Cartesian coordinates
http://web.physics.ucsb.edu/~fratus/phys103/Disc/disc_notes_3_pdf.pdf
2. http://planetmath.org/unitvectorsincurvilinearcoordinates to know more about curvilinear cordinates
3. Find extra stuff at this link http://mathworld.wolfram.com/SphericalCoordinates.html

Definitely there is much much more information to be added to this topic. I’ll do it some other time.